Question

If the two diagonals of one of its faces are $6\hat{i}+6\hat{k}$ and $4\hat{j}+2\hat{k}$ and of the edges not containing the given diagonals is $\overrightarrow{c}=4\hat{i}-8\hat{k}$, then the volume of a parallelepiped is

• A. $60$
• B. $80$
• C. $100$
• D. $120$

Answer 1

The correct option is D $120$

Let $\overrightarrow{a}=6\hat{i}+6\hat{k},\overrightarrow{b}=4\hat{j}+2\hat{k},\overrightarrow{c}=4\hat{j}-8\hat{k}$

Then $\overrightarrow{a}\times \overrightarrow{b}=-24\hat{i}-12\hat{j}+24\hat{k}=12(-2\hat{i}-\hat{j}+2\hat{k})$
$\therefore$ Area of the base of the parallelepiped

$=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$
$=\frac{1}{2}(12\times 3)$
$=18$

Height of the parallelepiped
$=$ Length of projection of $\overrightarrow{c}$ on $\overrightarrow{a}\times \overrightarrow{b}$
$=\frac{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}{|\overrightarrow{a}\times \overrightarrow{b}|}$
$=\frac{|12(-4-16)|}{36}$
$=\frac{20}{3}$
$\therefore$ Volume of the parallelepiped $=18\times \frac{20}{3}=120$